Casting pointer to member

[cbull]

class A
{
};
class B: public A
{
void f();
};

typedef void (A::*MPA)(void);
MPA mpA;

B b;
mpA = (MPA) &B::f;

b->*mpA();


Will it work? If not then when it will fail? I've heard something about
multi inheritance.

Tomasz

 

[Victor Bazarov]

class A
{
};
class B: public A
{
void f();
};

typedef void (A::*MPA)(void);
MPA mpA;

B b;
mpA = (MPA) &B::f;

b->*mpA();


Will it work? If not then when it will fail? I've heard something about
multi inheritance.

No, it won't work. B::f is not a member of A. Pretending that it is will
not help solving whatever problem you're trying to solve. BTW, what is it
that you're trying to accomplish?

V

 

[cbull]

No, it won't work. B::f is not a member of A. Pretending that it is will
not help solving whatever problem you're trying to solve. BTW, what is it
that you're trying to accomplish?

But I'm sure that I will always call it on correct object. And that is my
question. Will it always work while calling on object that this method
belongs to?
It's a part of slot - signal system.

Tomasz

 

[Victor Bazarov]

[..] Will it always work while calling on object that this method
belongs to?
It's a part of slot - signal system.

Actually, I am not sure. You'd need to dig into pointer adjustments
made to the member pointer when accessing (calling) a member of the base
for the object of the derived. Generally speaking, a pointer to member
function probably doesn't change when you cast it (or even when you
convert it properly). However, internally it may end up accessing the
wrong portion of the derived object.

V

 

[Heinz Ozwirk]

class A
{
};
class B: public A
{
void f();
};

typedef void (A::*MPA)(void);
MPA mpA;

B b;
mpA = (MPA) &B::f;

b->*mpA();


Will it work? If not then when it will fail? I've heard something about
multi inheritance.

First I thought, such things cannot work, but after reading some pages of
the standard I have my doubts

4.11 (Standard Conversions) says "An rvalue of type "pointer to member of B
of type cv T," where B is a class type, can be converted to an rvalue of
type "pointer to member of D of type cv T," where D is a derived class
(clause 10) of B."

Later 5.2.9 (static_cast) says "An rvalue of type "pointer to member of D of
type cv1 T" can be converted to an rvalue of type "pointer to member of B of
type cv2 T", where B is a base class (clause 10) of D, if a valid standard
conversion from "pointer to member of B of type T" to "pointer to member of
D of type T" exists (4.11), and cv2 is the same cv-qualification as, or
greater cv-qualification than, cv1"

Putting those pieces together I come to the conclusion that it should work,
if I correctly understand those parts of the standards and didn't miss other
important parts (for this problem). According to 4.11 there is a standard
conversion from void(A::*)() [aka MPA] to void(B::*)(), and, according to
5.2.9, if there is such a conversion, you can explicitly cast from void
(B::*)() to void (A::*)() using a static_cast.

HTH
Heinz

 

[Victor Bazarov]

class A
{
};
class B: public A
{
void f();
};

typedef void (A::*MPA)(void);
MPA mpA;

B b;
mpA = (MPA) &B::f;

b->*mpA();


Will it work? If not then when it will fail? I've heard something about
multi inheritance.

[...] I come to the conclusion that it should work,
if I correctly understand those parts of the standards and didn't miss other
important parts (for this problem). According to 4.11 there is a standard
conversion from void(A::*)() [aka MPA] to void(B::*)(), and, according to
5.2.9, if there is such a conversion, you can explicitly cast from void
(B::*)() to void (A::*)() using a static_cast.

One thing is to convert pointers to members that way. Using them seems to
me a whole different matter.

V

 

[cbull]

class A
{
};
class B: public A
{
void f();
};

typedef void (A::*MPA)(void);
MPA mpA;

B b;
mpA = (MPA) &B::f;

b->*mpA();


Will it work? If not then when it will fail? I've heard something about
multi inheritance.

Tomasz

I've tried this construction under VC++ 2005 and it seems to work. Later I
will try it under gcc.

Tomasz